IPMAT Indore 2020 MCQ PYQs

Q1:ipmat indore 2020QA › ProbabilityHardMCQ · MCQ

The probability that a randomly chosen factor of

10^{19}

is a multiple of

10^{15}

is

A $\frac{1}{25}$
B $\frac{1}{12}$
C $\frac{1}{20}$
D $\frac{1}{16}$

Pick an option to attempt

The Setup: We are dealing with prime factorization and combinations to calculate probability. We need to find the total number of factors of

10^{19}

(the sample space) and the total number of factors that carry at least

10^{15}

in their DNA (the winning outcomes). Step 1: Break down the base stats. Prime factorize the main number:

10^{19}=(2\times 5)^{19}=2^{19}\times 5^{19}

The formula for the total number of factors of

p^a\times q^b

is

(a+1)(b+1)

, since each exponent is chosen freely:

\text{Total factors}=(19+1)(19+1)=20\times 20=400

Step 2: Isolate the target condition. A factor of

10^{19}

has the shape

2^x\times 5^y

with

0\leq x,y\leq 19

. For it to be a multiple of

10^{15}=2^{15}\times 5^{15}

, it must carry at least fifteen of each prime, so

x\geq 15

and

y\geq 15

. Combined with the ceiling of 19, the valid range for both

x

and

y

is

\{15,16,17,18,19\}

. Step 3: Count the winning outcomes. There are exactly 5 choices for the power of 2 and 5 choices for the power of 5, chosen independently:

\text{Valid factors}=5\times 5=25

Step 4: Calculate the final RNG. Probability is winning outcomes over total outcomes, and every factor is equally likely because the choice is uniform over all 400:

P=\frac{25}{400}=\frac{1}{16}

Final Answer:

\frac{1}{16}

Q2:ipmat indore 2020QA › TrianglesHardMCQ · MCQ

The number of acute angled triangles whose sides are three consecutive positive integers and whose perimeter is at most 100 is

A28
B29
C31
D33

Pick an option to attempt

The Setup: This geometry problem is a triple filter. The sides must pass the triangle inequality vibe check, then survive the acute-angle condition, then fit under the perimeter cap. Whichever filter is strictest wins. Step 1: Set up the side lengths. Let the three consecutive integer sides be

x-1

,

x

and

x+1

, with

x-1\geq 1

. For any triangle the two smaller sides must beat the largest:

(x-1)+x>x+1\implies x>2

. Step 2: Enforce the acute angle rule. A triangle is acute when *every* angle is acute - but we only have to test one. The largest angle always sits opposite the longest side, so if that angle is acute the two smaller ones are smaller still and acute automatically. The longest side here is

x+1

, so the single condition to check is

a^2+b^2>c^2

:

(x-1)^2+x^2>(x+1)^2

x^2-2x+1+x^2>x^2+2x+1

x^2>4x

Since

x

is positive we may divide by

x

, giving

x>4

. As

x

is an integer, the lowest valid value is

x=5

(the smallest qualifying triangle being

4,5,6

). Note this condition is strictly tighter than Step 1's

x>2

, so the triangle inequality is absorbed and adds nothing further. Step 3: Apply the perimeter cap. The perimeter is at most 100:

(x-1)+x+(x+1)=3x\leq 100 \implies x\leq 33.33\ldots

Since

x

is an integer, the maximum is

x=33

, giving sides

32,33,34

and a perimeter of

99

. Step 4: Count the surviving squad. The valid range for

x

runs from 5 to 33 inclusive:

33-5+1=29

Final Answer: 29

Q3:ipmat indore 2020QA › Straight LinesMediumMCQ · MCQ

The equation of the straight line passing through the point

M (-5,4)

, such that the portion of it between the axes is divided by the point

M

into two equal halves, is

A $10y - 8x = 80$
B $8y + 10x = 80$
C $10y + 8x = 80$
D $8y + 10x + 80 = 0$

Pick an option to attempt

The Setup: This is a coordinate geometry speedrun using intercept form. The line crosses the x-axis at some point

A(a,0)

and the y-axis at

B(0,b)

. Saying

M

divides the portion between the axes into two equal halves is just a fancy way of saying

M

is the midpoint of segment

AB

. Step 1: Apply the midpoint formula. The midpoint of

A(a,0)

and

B(0,b)

is

\left(\frac{a}{2},\frac{b}{2}\right)

, and we are told that point is

M(-5,4)

:

\frac{a}{2}=-5\implies a=-10

\frac{b}{2}=4\implies b=8

Step 2: Construct the line equation. Use the two-intercept form

\frac{x}{a}+\frac{y}{b}=1

:

\frac{x}{-10}+\frac{y}{8}=1

Step 3: Clear the fractions in one shot. Multiply every term by

80

(a common multiple of 10 and 8 chosen to land directly on the answer's scale):

80\cdot\frac{x}{-10}+80\cdot\frac{y}{8}=80 \implies -8x+10y=80

Reordering gives

10y-8x=80

. Step 4: Verify against all three known points. A midpoint slip is easy to make, so test the line on the two intercepts and

M

itself: *

A(-10,0)

:

10(0)-8(-10)=80

✓ *

B(0,8)

:

10(8)-8(0)=80

✓ *

M(-5,4)

:

10(4)-8(-5)=40+40=80

✓ All three sit on the line, and

M

is genuinely halfway between

A

and

B

. Running the other three options through

M(-5,4)

fails every time, so the match is unique. Final Answer:

10y - 8x = 80

Q4:ipmat indore 2020QA › TrigonometryHardMCQ · MCQ

The value of

\cos^2\left(\frac{\pi}{8}\right) + \cos^2\left(\frac{3\pi}{8}\right) + \cos^2\left(\frac{5\pi}{8}\right) + \cos^2\left(\frac{7\pi}{8}\right)

is

A $1$
B $\frac{3}{2}$
C $2$
D $\frac{9}{4}$

Pick an option to attempt

The Setup: This trig expression looks intimidating, but it has zero plot armour once you spot the symmetry in the angles. We use the supplementary angle identity to fold the expression in half, then the complementary identity to finish it off. Step 1: Fold the back half. Notice that

\frac{7\pi}{8}=\pi-\frac{\pi}{8}

, and

\cos(\pi-\theta)=-\cos\theta

, so

\cos\left(\frac{7\pi}{8}\right)=-\cos\left(\frac{\pi}{8}\right)

. Squaring kills the minus sign:

\cos^2\left(\frac{7\pi}{8}\right)=\cos^2\left(\frac{\pi}{8}\right)

Identically,

\frac{5\pi}{8}=\pi-\frac{3\pi}{8}

, so

\cos^2\left(\frac{5\pi}{8}\right)=\cos^2\left(\frac{3\pi}{8}\right)

. Step 2: Simplify the sum. The four terms collapse into two identical pairs:

2\left[\cos^2\left(\frac{\pi}{8}\right)+\cos^2\left(\frac{3\pi}{8}\right)\right]

Step 3: Exploit the complementary angles. Notice that

\frac{\pi}{8}+\frac{3\pi}{8}=\frac{4\pi}{8}=\frac{\pi}{2}

. The two angles are complementary, and

\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta

, so:

\cos\left(\frac{3\pi}{8}\right)=\sin\left(\frac{\pi}{8}\right)

Step 4: Use the legendary identity. Substitute sine into the folded expression:

2\left[\cos^2\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{8}\right)\right]

By the Pythagorean identity

\cos^2\theta+\sin^2\theta=1

, the bracket collapses to 1:

2\times 1=2

Final Answer:

2

Q5:ipmat indore 2020QA › Progression & SeriesHardMCQ · MCQ

If

\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \ldots

up to

\infty = \frac{\pi^2}{6}

, then the value of

\frac{1}{1^2} + \frac{1}{3^2} + \frac{1}{5^2} + \ldots

up to

\infty

is

A $\frac{\pi^2}{8}$
B $\frac{\pi^2}{16}$
C $\frac{\pi^2}{12}$
D $\frac{\pi^2}{36}$

Pick an option to attempt

The Setup: We are manipulating the famous Basel problem. We are handed the sum of *all* inverse squares and asked for the sum of only the *odd* ones. The trick is to split the total into its even and odd halves, then notice the even half is a scaled copy of the whole thing. Step 1: Define the series. Let the total be

S=\frac{\pi^2}{6}

. Every positive integer is either odd or even, so the total splits cleanly with nothing counted twice and nothing missed:

S=\text{Sum(Odds)}+\text{Sum(Evens)}

Every term here is positive and the series converges, so splitting and rearranging it like this is legitimate rather than a formal trick. Step 2: Factor the even series. Write out the even terms and pull out the common factor. Each even number is

2n

, so its square is

4n^2

:

\text{Sum(Evens)}=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\ldots=\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots\right)

The bracket is literally the original total

S

, so

\text{Sum(Evens)}=\frac{S}{4}

. This self-similarity is the whole engine of the problem. Step 3: Solve for the odds. Substitute back into the master equation:

S=\text{Sum(Odds)}+\frac{S}{4} \implies \text{Sum(Odds)}=S-\frac{S}{4}=\frac{3S}{4}

Step 4: Final substitution. Plug in the known value of

S

:

\text{Sum(Odds)}=\frac{3}{4}\times \frac{\pi^2}{6}=\frac{3\pi^2}{24}=\frac{\pi^2}{8}

Final Answer:

\frac{\pi^2}{8}

Q6:ipmat indore 2020QA › ProbabilityHardMCQ · MCQ

A man is known to speak the truth on an average 4 out of 5 times. He throws a die and reports that it is a five. The probability that it is actually a five is

A $\frac{4}{9}$
B $\frac{5}{9}$
C $\frac{4}{15}$
D $\frac{2}{15}$

Pick an option to attempt

The Setup: This is a classic Bayes' Theorem trap. We have a prior probability (the physical die) mixed with a conditional one (the man's honesty stats). We want the probability that a five *actually* came up, given that he claims it did - which is not the same as his truthfulness rate, and the gap between

\frac{4}{5}

and the answer is the entire point of the question. Step 1: Map the base probabilities. * Probability of rolling a five:

P(A)=\frac{1}{6}

* Probability of rolling anything else:

P(\text{not }A)=\frac{5}{6}

* Man tells the truth:

P(T)=\frac{4}{5}

* Man lies:

P(L)=\frac{1}{5}

Step 2: Map the conditional claims. If it genuinely is a five, he reports five precisely when he is being truthful:

P(\text{Claims 5}\mid \text{Is 5})=\frac{4}{5}

. If it is not a five, he can only report five by lying, so this branch is taken as

P(\text{Claims 5}\mid \text{Not 5})=\frac{1}{5}

- the exam's convention is that a lie about the target face lands on that face. *Worth naming the assumption, because it is doing real work.* If instead a liar spread his lie uniformly across the five wrong faces, this branch would be

\frac{1}{5}\times\frac{1}{5}=\frac{1}{25}

and the answer would come out as

\frac{4}{5}

. That value appears nowhere in the options, whereas the convention above lands exactly on option 1 - so the option set itself tells us which reading the paper intended. Step 3: Apply Bayes' Theorem.

P(\text{Is 5}\mid \text{Claims 5})=\frac{P(\text{Is 5})\times P(\text{Claims 5}\mid \text{Is 5})}{P(\text{Is 5})\times P(\text{Claims 5}\mid \text{Is 5})+P(\text{Not 5})\times P(\text{Claims 5}\mid \text{Not 5})}

Step 4: Crunch the numbers.

\text{Numerator}=\frac{1}{6}\times \frac{4}{5}=\frac{4}{30}

\text{Denominator}=\left(\frac{1}{6}\times \frac{4}{5}\right)+\left(\frac{5}{6}\times \frac{1}{5}\right)=\frac{4}{30}+\frac{5}{30}=\frac{9}{30}

P=\frac{\ \frac{4}{30}\ }{\frac{9}{30}}=\frac{4}{9}

Sanity check on the direction:

\frac{4}{9}<\frac{1}{2}

, which is right - a five is rare enough on a fair die that even a mostly-honest report leaves it more likely than not that he is lying. Final Answer:

\frac{4}{9}

Q7:ipmat indore 2020QA › LogarithmsEasyMCQ · MCQ

If

\log_5(\log_8(x^2 - 1)) = 0

, then a possible value of

x

is

A2 $\sqrt{2}$
B $\sqrt{2}$
C2
D3

Pick an option to attempt

The Setup: This is a logarithm inception. We unpack it from the outside in using the fundamental definition

\log_b(A)=C\implies A=b^C

, then check the candidates against the option list. Step 1: Destroy the outer log.

\log_5\left(\log_8(x^2-1)\right)=0

Convert from log form to exponential form:

\log_8(x^2-1)=5^0=1

Step 2: Destroy the inner log. Apply the same rule to what remains:

x^2-1=8^1=8

Step 3: Solve the quadratic.

x^2=9 \implies x=\pm 3

Step 4: Pick the value that is actually on the menu. Both

x=3

and

x=-3

satisfy the equation - the expression only ever sees

x^2

, so the sign is irrelevant, and

x^2-1=8>0

keeps every logarithm well defined. This is exactly why the stem asks for *a possible* value rather than *the* value. Of the two, only

3

appears among the options, so it is the answer. The other three options all fail, and one of them is a genuine trap: *

x=2\sqrt{2}

:

x^2-1=7

, and

\log_8 7\approx 0.936

, so the outer log is not 0. *

x=\sqrt{2}

:

x^2-1=1

, so

\log_8 1=0

- and

\log_5 0

is undefined, not zero. Mistaking one for the other is the intended snare. *

x=2

:

x^2-1=3

, and

\log_8 3\approx 0.528

, so the outer log is not 0. Final Answer: 3

Q8:ipmat indore 2020QA › InequalitiesMediumMCQ · MCQ

Consider the following statements: (i) When

(0 < x < 1)

, then

(\frac{1}{1+x} < 1 - x + x^2)

(ii) When

(0 < x < 1)

, then

(\frac{1}{1+x} > 1 - x + x^2)

(iii) When

(-1 < x < 0)

, then

(\frac{1}{1+x} < 1 - x + x^2)

(iv) When

(-1 < x < 0)

, then

(\frac{1}{1+x} > 1 - x + x^2)

Then the correct statements are:

A(i) and (ii)
B(ii) and (iv)
C(i) and (iv)
D(ii) and (iii)

Pick an option to attempt

The Setup: Cross-multiplying an inequality whose denominator changes sign is how this question eats people alive. We dodge that entirely with the sum of cubes identity

a^3+b^3=(a+b)(a^2-ab+b^2)

, taking

a=1

and

b=x

, which turns the whole comparison into a single fraction whose sign we can read off. Step 0: Note the structure. Statements (i) and (ii) are exact opposites on the same interval, as are (iii) and (iv). So exactly one of each pair must be true, and the answer is guaranteed to be one statement from each pair - that alone kills option 1 ((i) and (ii) cannot both hold) before any algebra. Step 1: Exploit the identity. With

a=1

,

b=x

:

1+x^3=(1+x)(1-x+x^2)

On both intervals

x>-1

, so

1+x>0

- it is never zero, and we may safely divide by it:

1-x+x^2=\frac{1+x^3}{1+x}

Step 2: Find the delta. Subtract the two quantities being compared and put them over the common denominator:

(1-x+x^2)-\frac{1}{1+x}=\frac{1+x^3}{1+x}-\frac{1}{1+x}=\frac{x^3}{1+x}

Everything now rides on the sign of

\frac{x^3}{1+x}

, and since

1+x>0

throughout, that sign is simply the sign of

x^3

- which is the sign of

x

. **Step 3: Evaluate Case 1,

0<x<1

.** The numerator

x^3

is positive and the denominator

1+x

is positive, so the delta is positive:

\frac{x^3}{1+x}>0 \implies 1-x+x^2>\frac{1}{1+x}

Flipping that around to the form the paper uses gives

\frac{1}{1+x}<1-x+x^2

, which is exactly Statement (i) - and therefore (ii) is false. *Spot check at

x=\frac{1}{2}

:*

\frac{1}{1+x}=\frac{2}{3}

and

1-x+x^2=\frac{3}{4}

, and indeed

\frac{2}{3}<\frac{3}{4}

, with the gap

\frac{1}{12}

matching

\frac{x^3}{1+x}=\frac{1/8}{3/2}=\frac{1}{12}

. **Step 4: Evaluate Case 2,

-1<x<0

.** Here the numerator

x^3

is negative (a negative number cubed stays negative), while the denominator

1+x

is still positive - for instance

1+(-0.5)=0.5

. A negative over a positive is negative:

\frac{x^3}{1+x}<0 \implies 1-x+x^2<\frac{1}{1+x}

Flipping it round gives

\frac{1}{1+x}>1-x+x^2

, which is exactly Statement (iv) - and therefore (iii) is false. *Spot check at

x=-\frac{1}{2}

:*

\frac{1}{1+x}=2

and

1-x+x^2=\frac{7}{4}

, and indeed

2>\frac{7}{4}

, with the gap

-\frac{1}{4}

matching

\frac{x^3}{1+x}=\frac{-1/8}{1/2}=-\frac{1}{4}

. Step 5: Assemble. The true statements are (i) and (iv), one from each contradictory pair as Step 0 predicted. Final Answer: (i) and (iv)

Q9:ipmat indore 2020QA › Mixture & AlligationMediumMCQ · MCQ

Fifty litres of a mixture of milk and water contains 30 percent of water. This mixture is added to eighty litres of another mixture of milk and water that contains 20 percent of water. Then, how many litres of water should be added to the resulting mixture to obtain a final mixture that contains 25 percent of water?

A1
B2
C3
D4

Pick an option to attempt

The Setup: Mixture problems just require tracking the absolute volumes of the components like inventory. We find the exact litres of water and milk we currently hold, merge them, then solve for the extra water needed to hit the target 25%. The key discipline is that adding pure water changes the water column *and* the total, but leaves the milk column frozen. Step 1: Audit Mixture 1. Total

=50

L, water is 30%.

\text{Water}=0.30\times 50=15\text{ L}, \qquad \text{Milk}=50-15=35\text{ L}

Step 2: Audit Mixture 2. Total

=80

L, water is 20%.

\text{Water}=0.20\times 80=16\text{ L}, \qquad \text{Milk}=80-16=64\text{ L}

Step 3: Merge the inventory. Tip both into the same barrel:

\text{Total volume}=50+80=130\text{ L}, \quad \text{Water}=15+16=31\text{ L}, \quad \text{Milk}=35+64=99\text{ L}

Step 4: Calculate the required additive. Let

w

litres of pure water go in. Both the water and the total grow by

w

, and a 25% water mix means water is

\frac{1}{4}

of the total:

\frac{31+w}{130+w}=\frac{1}{4}

Cross-multiply and solve:

4(31+w)=130+w \implies 124+4w=130+w \implies 3w=6 \implies w=2

Step 5: Verify from the milk side. The milk column never moved, so it is an independent check. Final total

=130+2=132

L, of which milk is still 99 L:

\frac{99}{132}=\frac{3}{4}=75\% \text{ milk}, \text{ hence } 25\% \text{ water}

The two routes agree, so

w=2

is right. Final Answer: 2

Q10:ipmat indore 2020QA › Time & WorkHardMCQ · MCQ

Three workers working together need 1 hour to construct a wall. The first worker, working alone, can construct the wall twice as fast as the third worker, and can complete the task an hour sooner than the second worker. Then, the average of time taken in hours by three workers, when working alone, to construct the wall is

A $\dfrac{\sqrt{33} + 4}{3}$
B $\dfrac{\sqrt{33} + 5}{3}$
C $\dfrac{\sqrt{33} + 6}{3}$
D $\dfrac{\sqrt{33} + 7}{3}$

Pick an option to attempt

The Setup: This is a rates problem, and the golden rule is that *times do not add, rates do*. We convert *hours to build the wall* into *walls per hour*, sum the three rates to 1 wall/hour, and only convert back to time at the very end. Step 1: Define the rates. Let

r_1,r_2,r_3

be the workers' rates in walls per hour, and

t_i=\frac{1}{r_i}

their solo times. All three together take 1 hour, so their rates sum to one whole wall per hour:

r_1+r_2+r_3=\frac{1}{1}=1

Worker 1 is twice as fast as Worker 3, so

r_1=2r_3

, i.e.

r_3=\frac{r_1}{2}

(equivalently

t_3=2t_1

). Step 2: Link Workers 1 and 2. Worker 1 finishes an hour sooner than Worker 2, and that condition is about *times*, so it must be written in times first:

t_1=t_2-1 \implies \frac{1}{r_1}=\frac{1}{r_2}-1 \implies \frac{1}{r_2}=\frac{1}{r_1}+1=\frac{1+r_1}{r_1}

Flipping gives

r_2=\frac{r_1}{1+r_1}

. Step 3: Solve the master rate equation. Substitute

r_2

and

r_3

into the sum:

r_1+\frac{r_1}{1+r_1}+\frac{r_1}{2}=1

Multiply through by

2(1+r_1)

to clear denominators:

2r_1(1+r_1)+2r_1+r_1(1+r_1)=2(1+r_1)

2r_1+2r_1^2+2r_1+r_1+r_1^2=2+2r_1

3r_1^2+5r_1=2+2r_1 \implies 3r_1^2+3r_1-2=0

By the quadratic formula:

r_1=\frac{-3\pm\sqrt{9-4(3)(-2)}}{6}=\frac{-3\pm\sqrt{33}}{6}

A rate cannot be negative, and

\sqrt{33}\approx 5.745

, so the minus root is discarded and

r_1=\frac{\sqrt{33}-3}{6}

. Step 4: Convert back to times. Rationalising the denominator each time:

t_1=\frac{1}{r_1}=\frac{6}{\sqrt{33}-3}=\frac{6(\sqrt{33}+3)}{33-9}=\frac{6(\sqrt{33}+3)}{24}=\frac{\sqrt{33}+3}{4}

t_2=t_1+1=\frac{\sqrt{33}+3}{4}+\frac{4}{4}=\frac{\sqrt{33}+7}{4}, \qquad t_3=2t_1=\frac{2\sqrt{33}+6}{4}

Step 5: Average them. Note the question asks for the average of the *times*, not of the rates - averaging rates instead is the single most common way to lose this mark.

t_1+t_2+t_3=\frac{(\sqrt{33}+3)+(\sqrt{33}+7)+(2\sqrt{33}+6)}{4}=\frac{4\sqrt{33}+16}{4}=\sqrt{33}+4

\text{Average}=\frac{\sqrt{33}+4}{3}

Step 6: Sanity check the numbers. Numerically

t_1\approx 2.19

,

t_2\approx 3.19

,

t_3\approx 4.37

hours. Every one exceeds 1 hour, as it must - a lone worker cannot beat all three together. Their gap

t_2-t_1

is exactly 1,

t_3

is exactly double

t_1

, and

\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}=1

to twelve decimal places. All three given conditions hold. Final Answer:

\dfrac{\sqrt{33} + 4}{3}

Q11:ipmat indore 2020QA › Set TheoryMediumMCQ · MCQ

In a class, students are assigned roll numbers from 1 to 140. All students with even roll numbers opted for cricket, all those whose roll numbers are divisible by 5 opted for football, and all those whose roll numbers are divisible by 3 opted for basketball. The number of students who did not opt for any of the three sports is

A102
B38
C98
D42

Pick an option to attempt

The Setup: This is Inclusion-Exclusion. We count the students playing at least one sport (the union of the three sets), then subtract from 140 to get the NPCs playing nothing. Note the sets overlap heavily - a roll number like 30 is in all three - which is exactly why we cannot simply add. Step 1: Calculate the single sets. Counting multiples of

k

up to 140 is

\left\lfloor\frac{140}{k}\right\rfloor

.

|A|\ (\text{cricket, }2)=\left\lfloor\frac{140}{2}\right\rfloor=70, \quad |B|\ (\text{football, }5)=\left\lfloor\frac{140}{5}\right\rfloor=28, \quad |C|\ (\text{basketball, }3)=\left\lfloor\frac{140}{3}\right\rfloor=46

Step 2: Calculate the pairwise intersections. A student in two sets has a roll number divisible by both divisors, so the correct counter is their LCM - not their product. Here

2,3,5

are pairwise coprime, so the LCM happens to equal the product, but that coincidence is what makes this step safe; with divisors like 2 and 4 the product would give a wrong count.

|A\cap B|\ (\text{LCM}=10)=\left\lfloor\frac{140}{10}\right\rfloor=14, \quad |A\cap C|\ (\text{LCM}=6)=\left\lfloor\frac{140}{6}\right\rfloor=23, \quad |B\cap C|\ (\text{LCM}=15)=\left\lfloor\frac{140}{15}\right\rfloor=9

Step 3: Calculate the triple intersection.

|A\cap B\cap C|\ (\text{LCM}=30)=\left\lfloor\frac{140}{30}\right\rfloor=4

Step 4: Execute the formula.

|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|

=70+28+46-14-23-9+4=144-46+4=102

Step 5: Take the complement. That 102 is the number who play at least one sport, so:

140-102=38

Watch the trap: 102 is sitting right there as option 1. It is the union, not the answer - the question asks for everyone *outside* it. Final Answer: 38

Q12:ipmat indore 2020QA › FunctionsEasyMCQ · MCQ

Given

f(x) = x^2 + \log_3 x

and

g(y) = 2y + f(y)

, then the value of

g(3)

equals

A16
B15
C25
D26

Pick an option to attempt

The Setup: Pure function evaluation. No theory required - just substitute from the outside in, respecting each definition exactly as written. Step 1: Set up the outer function. The definition is

g(y)=2y+f(y)

, so at

y=3

:

g(3)=2(3)+f(3)=6+f(3)

Note the variable name is a red herring:

f

is written in

x

and

g

in

y

, but a function does not care what its argument is called. Feeding 3 into

f

is the same operation either way. Step 2: Evaluate the nested function. Using

f(x)=x^2+\log_3 x

at

x=3

:

f(3)=3^2+\log_3 3

We have

3^2=9

, and any log of its own base is 1, since

\log_b b=1

means *what power of

b

gives

b

?* - namely 1. So

\log_3 3=1

and

x=3>0

keeps the log defined.

f(3)=9+1=10

Step 3: Recombine. Substitute back:

g(3)=6+10=16

Final Answer: 16

Q13:ipmat indore 2020QA › Matrices & DeterminantsHardMCQ · MCQ

A

2 \times 2

matrix is filled with four distinct integers randomly chosen from the set

\{1,2,3,4,5,6\}

. Then the probability that the matrix generated in such a way is singular is

A $\frac{2}{45}$
B $\frac{1}{45}$
C $\frac{4}{15}$
D $\frac{1}{15}$

Pick an option to attempt

The Setup: A

2\times 2

matrix

M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}

is singular when its determinant vanishes, i.e.

ad-bc=0

, so we need the main-diagonal product to equal the anti-diagonal product:

ad=bc

. The whole question reduces to finding which four numbers from

\{1,2,3,4,5,6\}

can be split into two disjoint pairs of equal product. Step 1: Calculate the total sample space. We pick 4 distinct numbers from 6 and place them in 4 labelled positions, so order matters - this is a permutation:

\text{Total outcomes}=P(6,4)=6\times 5\times 4\times 3=360

Step 2: Find every pair of pairs with equal products. Rather than assert it, list all 15 pairs and their products outright:

Pair	Product	Pair	Product	Pair	Product
(1,2)	2	(2,4)	8	(3,5)	15
(1,3)	3	(2,5)	10	(3,6)	18
(1,4)	4	(2,6)	12	(4,5)	20
(1,5)	5	(3,4)	12	(4,6)	24
(1,6)	6	(2,3)	6	(5,6)	30

Exactly two product values repeat, and in both cases the two pairs are disjoint (they share no element, which is required since all four entries must be distinct): * Product

6

:

\{1,6\}

and

\{2,3\}

* Product

12

:

\{2,6\}

and

\{3,4\}

Every other product in the table is unique, so no further singular matrices exist. Step 3: Count matrix arrangements. Take product 6. Put

\{1,6\}

on the main diagonal

(a,d)

and

\{2,3\}

on the anti-diagonal

(b,c)

: * Arrangements of

\{1,6\}

across

a,d

:

2!=2

* Arrangements of

\{2,3\}

across

b,c

:

2!=2

That gives

2\times 2=4

matrices. But the two pairs can also switch diagonals -

\{2,3\}

on the main,

\{1,6\}

on the anti - for another 4. So product 6 contributes

8

matrices, and by identical logic so does product 12. Step 4: Final probability.

\text{Singular matrices}=8+8=16 \implies P=\frac{16}{360}=\frac{2}{45}

Final Answer:

\frac{2}{45}

Q14:ipmat indore 2020QA › Ratio, Proportion & VariationEasyMCQ · MCQ

Ashok started a business with a certain investment. After a few months, Bharat joined him investing half the amount of Ashok's initial investment. At the end of the first year, the total profit was divided between them in the ratio

3:1

. Bharat joined Ashok after

A2 months
B3 months
C4 months
D6 months

Pick an option to attempt

The Setup: In partnership accounting, profit is shared in proportion to each partner's *capital-months* - the money put in multiplied by how long it stayed in. Two partners with the same capital but different durations do not split evenly, and that is the whole mechanism here. Step 1: Define Ashok's stats. Let Ashok's investment be

2x

(choosing

2x

rather than

x

keeps Bharat's half a whole number). He started the business, so his money was in for the full 12 months:

\text{Ashok's capital-months}=2x\times 12=24x

Step 2: Define Bharat's stats. Bharat invested half of Ashok's amount, so his capital is

x

. Let it stay in for

t

months:

\text{Bharat's capital-months}=x\times t=xt

Step 3: Equate to the profit ratio. Profit splits

3:1

in the order the partners are named, Ashok to Bharat:

\frac{\text{Ashok's capital-months}}{\text{Bharat's capital-months}}=\frac{3}{1} \implies \frac{24x}{xt}=3

Step 4: Solve for the duration. Capital is positive, so

x

cancels safely:

\frac{24}{t}=3 \implies 3t=24 \implies t=8

Step 5: Convert duration into a joining date - do not stop at Step 4. The number 8 is how long Bharat's money was *invested*, but the question asks when he joined. In a 12-month year, being invested for the last 8 months means he sat out the first four:

12-8=4\text{ months}

Answering 8 here is the intended trap, and it is why the question is phrased 'Bharat joined Ashok after' rather than asking for his investment period. Final Answer: 4 months

Q15:ipmat indore 2020QA › Mean, Median & ModeHardMCQ · MCQ

The average marks of

6

students in a test is

64

. All the students got different marks, one of the students obtained

70

marks and all other students scored

40

or above. The maximum possible difference between the second highest and the second lowest marks is

A50
B54
C57
D58

Pick an option to attempt

The Setup: A min-maxing puzzle wearing a statistics costume. Sort the six marks as

S_1>S_2>S_3>S_4>S_5>S_6

, so the second highest is

S_2

and the second lowest is

S_5

. We want to maximise

S_2-S_5

subject to a fixed total, all marks distinct integers of at least 40, and one of them being exactly 70. Step 1: Set the baseline constraints. Six students averaging 64 gives a fixed budget:

\text{Total}=6\times 64=384

Step 2: Pin the floor.

S_6\geq 40

, and since the marks are distinct integers,

S_5\geq S_6+1\geq 41

. So

S_5=41

is the best we can ever do, achieved by putting

S_6=40

. Every point saved at the bottom is a point available at the top, so this choice never costs us anything. Step 3: Decide where the mandatory 70 goes. This is the real work, and it needs all cases checked rather than a lucky guess: * **70 as

S_1

or

S_2

:** then

S_2\leq 70

, so the difference is at most

70-41=29

. Dead. * **70 as

S_5

or

S_6

:** impossible. If

S_5=70

then

S_1,\dots,S_4

all exceed 70, so they total at least

71+72+73+74=290

; adding

70

and

S_6\geq 40

already gives

400>384

. Placing 70 even lower is worse still. * **70 as

S_4

:** then

S_3\geq 71

, and with

S_5=41,S_6=40

the top three share

384-70-41-40=233

. Minimising

S_3

at 71 leaves

S_1+S_2=162

, and

S_1>S_2

forces

S_2\leq 80

. Difference at most

80-41=39

. Still beaten. * **70 as

S_3

: the surviving case, examined next. Step 4: Work the winning case.** With 70 at

S_3

, the two marks below it must be distinct integers between 41 and 69, and pushing them as low as possible frees the most budget for the top:

S_4=42

,

S_5=41

,

S_6=40

. That leaves:

S_1+S_2=384-(70+42+41+40)=384-193=191

To make

S_2

as large as possible we want

S_1

as close above it as possible. Since 191 is odd, the tightest legal split of

S_1>S_2

is

96

and

95

:

S_1=96,\quad S_2=95

Step 5: Confirm the set is legal and compute. The marks are

96,95,70,42,41,40

- six distinct integers, all at least 40, one exactly 70, summing to

384

. Every condition holds.

S_2-S_5=95-41=54

Since Step 3 ruled out every other placement of 70 with a strictly smaller ceiling, 54 is the true maximum. Final Answer: 54

Q16:ipmat indore 2020LRDI › Tabular DataMediumMCQ · MCQ

The table below presents the quoted buy and sell prices of five stocks during the five trading days of a given week. The quoted sell price is the price at which an investor can sell a stock in the market. The quoted buy price is the price at which an investor can buy a stock from the market. All the quoted numbers are in Indian Rupees.

Day	Monday	Tuesday	Wednesday	Thursday	Friday
Quote	Sell	Buy	Sell	Buy	Sell	Buy	Sell	Buy	Sell	Buy
Dabur	460	462	455	458	432	433	444	447	461	462
Marico	345	346	335	336	365	368	372	375	372	374
HUL	1931	1933	1952	1955	1979	1981	2044	2048	1966	1969
ITC	237	238	238	239	246	251	221	225	253	256
Britannia	3044	3046	3100	3101	3110	3115	3025	3027	3140	3144

show less

If an investor had Rs 36,00,000 to invest in any particular single stock, and she could buy the stock only on Monday and sell it off only on Friday, then the stock she should buy on Monday to earn the maximum possible profit during the week is

AMarico
BHUL
CITC
DBritannia

Pick an option to attempt

The Setup: Every question in this set runs on one rule, so it is worth nailing down once here. The table gives two prices per day, and the passage defines them from the *investor's* side: you buy at the Buy price and sell at the Sell price. The Buy price is always the higher of the two, so the market takes a small bite on every round trip. Mixing the two columns up is the single biggest source of wrong answers in this caselet. Step 1: Fix the method. With a fixed pot of money you cannot own a fraction of a share, so:

\text{Shares}=\left\lfloor\frac{\text{Capital}}{\text{Monday Buy price}}\right\rfloor, \qquad \text{Residual cash}=\text{Capital}-\text{Shares}\times\text{Buy price}

\text{Final wealth}=\text{Shares}\times\text{Friday Sell price}+\text{Residual cash}

The residual is the small change that could not buy one more share. It does not vanish - it stays in the investor's pocket and counts toward the final wealth. It barely matters here, but in MCQ 17 it decides the answer outright, so build the habit now. Step 2: Run all four candidates. Capital is Rs 36,00,000. Buy at Monday's Buy price, sell at Friday's Sell price.

Stock	Mon Buy	Shares	Residual	Fri Sell	Proceeds	Profit
Marico	346	10,404	216	372	38,70,288	2,70,504
HUL	1,933	1,862	754	1,966	36,60,692	61,446
ITC	238	15,126	12	253	38,26,878	2,26,890
Britannia	3,046	1,181	2,674	3,140	37,08,340	1,11,014

Two share counts are worth checking by hand, because rounding the wrong way is easy here:

3046\times 1181=35,97,326\leq 36,00,000 \quad\text{but}\quad 3046\times 1182=36,00,372>36,00,000

so Britannia buys 1,181 shares, not 1,182. Likewise

346\times 10404=35,99,784

fits while

346\times 10405=36,00,130

does not. Step 3: Read off the winner. Marico returns the largest profit at roughly Rs 2.70 lakh, comfortably ahead of ITC's Rs 2.27 lakh. Step 4: Sanity-check why. Marico wins on percentage movement, not on price. It rose from 346 to 372, a gain of about

7.5\%

, versus ITC's

6.3\%

, Britannia's

3.1\%

and HUL's

1.7\%

. Because the whole pot goes into one stock either way, the cheapest share price is irrelevant - only the percentage gain matters, and a quick ratio scan of

\frac{\text{Fri Sell}}{\text{Mon Buy}}

would have identified Marico without computing a single share count. Final Answer: Marico

Q17:ipmat indore 2020LRDI › Tabular DataMediumMCQ · MCQ

The table below presents the quoted buy and sell prices of five stocks during the five trading days of a given week. The quoted sell price is the price at which an investor can sell a stock in the market. The quoted buy price is the price at which an investor can buy a stock from the market. All the quoted numbers are in Indian Rupees.

Day	Monday	Tuesday	Wednesday	Thursday	Friday
Quote	Sell	Buy	Sell	Buy	Sell	Buy	Sell	Buy	Sell	Buy
Dabur	460	462	455	458	432	433	444	447	461	462
Marico	345	346	335	336	365	368	372	375	372	374
HUL	1931	1933	1952	1955	1979	1981	2044	2048	1966	1969
ITC	237	238	238	239	246	251	221	225	253	256
Britannia	3044	3046	3100	3101	3110	3115	3025	3027	3140	3144

show less

If an investor planned to invest Rs 36,00,000 in purchasing the stocks of HUL on Monday, sell them off on Wednesday and use the entire proceeds to purchase the stocks of Britannia on the same day and sell them off again on Friday, then the total investment return during the week would be

A2.80 percent
B3.00 percent
C3.20 percent
D3.40 percent

Pick an option to attempt

The Setup: A two-leg trade: HUL from Monday to Wednesday, then the entire proceeds into Britannia from Wednesday to Friday. Using the convention from MCQ 16 - buy at the Buy price, sell at the Sell price, and carry the residual cash forward. That last point is not bookkeeping pedantry here; it is what separates two of the four options. Step 1: Leg one - HUL, Monday to Wednesday. Monday Buy price is 1,933:

\text{Shares}=\left\lfloor\frac{36,00,000}{1933}\right\rfloor=1862, \qquad \text{Residual}=36,00,000-1862\times 1933=754

Sell on Wednesday at HUL's Sell price of 1,979:

1862\times 1979=36,84,898 \implies \text{Total cash}=36,84,898+754=36,85,652

Step 2: Leg two - Britannia, Wednesday to Friday. Wednesday Buy price is 3,115:

\left\lfloor\frac{36,85,652}{3115}\right\rfloor=1183 \quad\text{since}\quad 3115\times 1183=36,85,045\leq 36,85,652

This is the hinge of the whole question. Had we thrown away the Rs 754 left over from the HUL leg, the available cash would be 36,84,898 - and

36,85,045>36,84,898

, so only 1,182 shares would be affordable. That stray Rs 754 is precisely what pays for the 1,183rd share, and that one share is worth about Rs 3,140 by Friday. Discarding it drops the answer to

3.10\%

and lands you on the wrong option. Step 3: Close the position. Sell on Friday at Britannia's Sell price of 3,140, and add the residual from this leg (

36,85,652-36,85,045=607

):

1183\times 3140=37,14,620 \implies \text{Final wealth}=37,14,620+607=37,15,227

Step 4: Compute the return.

\text{Return}=\frac{37,15,227-36,00,000}{36,00,000}\times 100=\frac{1,15,227}{36,00,000}\times 100=3.2008\%\approx 3.20\%

Step 5: Confirm with a shortcut. Ignore the share-counting entirely and just compound the two price ratios, which is what the trade does at heart:

\frac{1979}{1933}\times\frac{3140}{3115}=1.03201 \implies 3.20\%

Both routes agree to two decimal places, which confirms the answer is robust and not an artefact of rounding. Final Answer: 3.20 percent

Q18:ipmat indore 2020LRDI › Tabular DataMediumMCQ · MCQ

The table below presents the quoted buy and sell prices of five stocks during the five trading days of a given week. The quoted sell price is the price at which an investor can sell a stock in the market. The quoted buy price is the price at which an investor can buy a stock from the market. All the quoted numbers are in Indian Rupees.

Day	Monday	Tuesday	Wednesday	Thursday	Friday
Quote	Sell	Buy	Sell	Buy	Sell	Buy	Sell	Buy	Sell	Buy
Dabur	460	462	455	458	432	433	444	447	461	462
Marico	345	346	335	336	365	368	372	375	372	374
HUL	1931	1933	1952	1955	1979	1981	2044	2048	1966	1969
ITC	237	238	238	239	246	251	221	225	253	256
Britannia	3044	3046	3100	3101	3110	3115	3025	3027	3140	3144

show less

The difference between the quoted buy and sell price of a stock is referred to as the spread of the stock. The average spread of the stocks is lowest on

AMonday
BTuesday
CThursday
DFriday

Pick an option to attempt

The Setup: The spread is defined in the stem as the gap between the quoted buy and sell price of a stock - the market's cut on a round trip. So for each day we take

\text{Buy}-\text{Sell}

for all five stocks and average the five numbers. Since every day has the same five stocks, comparing the totals is enough and the division by 5 is only cosmetic. Step 1: Compute the spread for each stock, day by day.

Day	Dabur	Marico	HUL	ITC	Britannia	Total	Average
Monday	2	1	2	1	2	8	1.6
Tuesday	3	1	3	1	1	9	1.8
Thursday	3	3	4	4	2	16	3.2
Friday	1	2	3	3	4	13	2.6

Worked out in full for Monday, the winning day:

\frac{(462-460)+(346-345)+(1933-1931)+(238-237)+(3046-3044)}{5}=\frac{2+1+2+1+2}{5}=\frac{8}{5}=1.6

Step 2: Compare. Monday's average of 1.6 is the lowest of the four days offered; Tuesday at 1.8 is the nearest rival. Step 3: Note the shortcut. Monday's total of 8 is the smallest total in the table, and no division was needed to see it. Monday is also the only day where three of the five stocks trade at a spread of just 1 or 2 across the board - it is the tightest, most liquid-looking day of the week. Final Answer: Monday

Q19:ipmat indore 2020LRDI › Tabular DataMediumMCQ · MCQ

The table below presents the quoted buy and sell prices of five stocks during the five trading days of a given week. The quoted sell price is the price at which an investor can sell a stock in the market. The quoted buy price is the price at which an investor can buy a stock from the market. All the quoted numbers are in Indian Rupees.

Day	Monday	Tuesday	Wednesday	Thursday	Friday
Quote	Sell	Buy	Sell	Buy	Sell	Buy	Sell	Buy	Sell	Buy
Dabur	460	462	455	458	432	433	444	447	461	462
Marico	345	346	335	336	365	368	372	375	372	374
HUL	1931	1933	1952	1955	1979	1981	2044	2048	1966	1969
ITC	237	238	238	239	246	251	221	225	253	256
Britannia	3044	3046	3100	3101	3110	3115	3025	3027	3140	3144

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A brokerage firm charges 0.1 percent trading commission on the value of shares bought or sold through its trading platform. If an investor bought 1000 shares of Britannia on Tuesday, and sold all of them on Thursday, then the total brokerage fee that will be charged from the investor is

A6,125
B6,126
C6,127
D6,128

Pick an option to attempt

The Setup: The commission is charged on the value of shares bought or sold, so it applies twice - once on the purchase leg and once on the sale leg - and each leg is valued at its own price. Using the convention from MCQ 16: the investor buys at Tuesday's Buy price and sells at Thursday's Sell price. Step 1: The buying leg, Tuesday. Britannia's Tuesday Buy price is 3,101:

\text{Value}=1000\times 3101=31,01,000 \implies \text{Fee}=31,01,000\times 0.1\%=3101

Step 2: The selling leg, Thursday. Britannia's Thursday Sell price is 3,025:

\text{Value}=1000\times 3025=30,25,000 \implies \text{Fee}=30,25,000\times 0.1\%=3025

Note that

0.1\%

of a value is just that value divided by 1,000 - and since exactly 1,000 shares were traded, each fee comes out equal to the share price itself. That is a pleasant shortcut, not a coincidence to rely on generally. Step 3: Total the two legs.

3101+3025=6126

Step 4: Read the option list - it is a trap laid with real precision. All four options are within Rs 3 of each other, because each one corresponds to a different mix-up of the Buy and Sell columns:

Tuesday price used	Thursday price used	Total	Option
Sell 3,100	Sell 3,025	6,125	1
Buy 3,101	Sell 3,025	6,126	2 - correct
Sell 3,100	Buy 3,027	6,127	3
Buy 3,101	Buy 3,027	6,128	4

There is no arithmetic in this question at all once you know which column to read; the entire difficulty is the convention. You buy at Buy and sell at Sell. Final Answer: 6,126

Q20:ipmat indore 2020LRDI › Tabular DataMediumMCQ · MCQ

The table below presents the quoted buy and sell prices of five stocks during the five trading days of a given week. The quoted sell price is the price at which an investor can sell a stock in the market. The quoted buy price is the price at which an investor can buy a stock from the market. All the quoted numbers are in Indian Rupees.

Day	Monday	Tuesday	Wednesday	Thursday	Friday
Quote	Sell	Buy	Sell	Buy	Sell	Buy	Sell	Buy	Sell	Buy
Dabur	460	462	455	458	432	433	444	447	461	462
Marico	345	346	335	336	365	368	372	375	372	374
HUL	1931	1933	1952	1955	1979	1981	2044	2048	1966	1969
ITC	237	238	238	239	246	251	221	225	253	256
Britannia	3044	3046	3100	3101	3110	3115	3025	3027	3140	3144

show less

If you had decided to invest Rs.36,00,000 worth of ITC stocks on Monday, then the day of the week you should choose to sell the stocks to earn the maximum possible profit would be

ATuesday
BWednesday
CThursday
DFriday

Pick an option to attempt

The Setup: The purchase is fixed - ITC bought on Monday - and only the exit day is ours to choose. Using the convention from MCQ 16: buy at Monday's Buy price of 238, sell at the chosen day's Sell price. Step 1: Note why no calculation is really needed. Monday's purchase locks in a fixed share count:

\left\lfloor\frac{36,00,000}{238}\right\rfloor=15,126 \text{ shares, with Rs 12 left over}

That count does not change with the exit day, and the residual is fixed too. So the final wealth is

15,126\times(\text{Sell price})+12

, which is strictly increasing in the sell price. Maximising profit is therefore identical to maximising ITC's Sell price, and the whole question collapses into scanning one row of the table. Step 2: Scan ITC's Sell prices.

Sell on	Tuesday	Wednesday	Thursday	Friday
ITC Sell price	238	246	221	253
Profit	0	1,21,008	-2,57,142	2,26,890

Step 3: Read off the answer. Friday's Sell price of 253 is the highest of the week, giving the maximum profit of about Rs 2.27 lakh. Thursday is the trap for anyone glancing at the Buy column or misreading the row - ITC dips to 221 that day, which would actually book a loss of roughly Rs 2.57 lakh against a purchase price of 238. Step 4: Sanity check. Selling on Tuesday at 238 exactly matches Monday's Buy price of 238, so that trade nets zero profit - a useful confirmation that the buy and sell columns are being read the right way round. Final Answer: Friday

IPMAT Indore 2020 — MCQ

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